as topic, thanks!
2007-09-28 20:58:58 · 5 answers · asked by vicehope 1 in Science & Mathematics ➔ Mathematics
.∫ t^2/t^3 + 4t/t^3 + 4/t^3 dx
=.∫ 1/t + 4t^-2 + 4t^-3 dx
=ln(t) - 4/t - 4t^-2/2 +C
=ln(t) - 4/t - 2/t^2 +C
sorry guys you all got it wrong!
2007-09-28 22:46:08 · answer #1 · answered by Anonymous · 2⤊ 1⤋
expand (t+2)^2/t^3 = (t^2 +4t +4)/t^3
then integrate term by term:
1/t dt + 4/t^2 dt + 4dt
integral of above= ln(t) - 4/t +t + constant
2007-09-28 21:05:41 · answer #2 · answered by tinkerson 1 · 1⤊ 2⤋
This is a simple integration already answered as above with a small typo error. Last term is 4t and not t.
Correct answer is ln(t) - 4/t +4t + constant
2007-09-28 21:39:40 · answer #3 · answered by Madhukar 7 · 0⤊ 2⤋
( t + 2 ) ² / t ³
= (t ² + 4 t + 4 ) / t ³
I = â« (1/t) + 4 t^(- 2) + 4 t^(- 3) dt
I = log t - 4 / t - 4 / t ² + C
2007-09-28 21:50:07 · answer #4 · answered by Como 7 · 0⤊ 2⤋
â« u^n du = u^n+1/(n+1) --- the first answer forgot to include the denominator.
2007-09-28 23:02:17 · answer #5 · answered by Shh! Be vewy, vewy quiet 6 · 0⤊ 0⤋